College-Test-Preparation-Princeton-Review-Cracking-the-SAT-Subject-Test-in-Chemistry-16th-Edition_-Everything-You-Need-to-Help-Score-a-Perfect-800-Princeton-Review-2017

26. C numbers. So the balanced equation becomes: 2C4H10(g) + 13O2(g) 27. D → 8CO2(g) + 10H2O(l). 28. E The mass of a proton is approximately 1 amu, and this is very 29. B nearly the mass of a neutron. Both a positron and an electron are 30. D much lighter than 1 amu. A hydrogen molecule weighs roughly twice as much as a proton, and an alpha particle weighs about four times as much. Add up the mass of 1 mole of this substance. From the periodic table, we know that • 1 mole of hydrogen atoms has a mass of about 1 g. • 1 mole of chlorine atoms has a mass of about 35 g. • 4 moles of oxygen atoms have a mass of about 64 g. The total gram-molecular weight of this substance, then, is 100 g. Oxygen’s contribution is 64 g, which means the compound is 64 percent oxygen by mass. As soon as you hear the term electron sharing, you know you’re dealing with a covalent bond, so eliminate (A), (B), and (C). The fact that the two atoms differ in electronegativity tells you that one has more attraction for the shared electrons than the other. The result? A polar covalent bond—the molecule has a negative and a positive pole. Use q = mc∆T to compute the amount of heat transfer. For water, the specific heat, c, is about 1 calorie/g • °C, so a 20 g sample of water experiencing a 20°C increase in temperature has (20 g)(1 calorie/g • °C)(20° C), or 400 calories of heat, transferred to it. Remember the oxidation state rules. An oxygen atom usually has a –2 state. Potassium atoms are always given a +1 state. In K2Cr2O7, there are 2 potassium atoms and 7 oxygen atoms. So potassium atoms contribute 2(+1), or a state of +2. Oxygen atoms contribute

31. B 2(–7), or –14. For K2Cr2O7 to be neutral, each chromium atom 32. D must have a state of +6. 33. B If the pH is 5, then [H+] is 1 × 10–5 moles/L. Water’s ion product is 34. C 1 × 10–14 at 25°C, meaning that the product [H+] × [OH–] is 1 × 35. C 10–14. So [OH–] = (1 × 10–14)/(1 × 10–5) = 1 × 10–9 moles/L. Choice (B) is correct. Based on the balanced equation, the ratio of water vapor consumed to oxygen produced is 2 moles H2O to 1 mole O2. The volume of gas will also be in this 2:1 ratio. So 89.6 liters of H2O(g) are required to produce 44.8 liters of O2(g). Molarity refers to moles of solute per liter of solution. You have 500 milliliters of solution, but you don’t know how many moles of solute you have. First figure out the mass of 1 mole of MgCl2. Looking at the periodic table, you can see that 1 mole of Mg has a mass of 24.3 g. Two moles of Cl have a mass of about 71 g. One mole of MgCl2, therefore, has a mass of 95.3 g. You can then determine the moles of MgCl2 by dividing the mass of the sample by the molar mass of the compound: 190 g/95.3 g = 2.0 mol. Finally, divide the moles of the sample by the volume of the solution (in liters) to determine the concentration of the solution: 2.0 mol/0.500 L = 4.0 M. The correct answer is (B). Think of the ideal gas equation: PV = nRT. What does this equation tell you? It means that volume is directly related to Kelvin temperature and inversely related to pressure. Doubling a gas’s Kelvin temperature will double its volume if other variables are held constant. Doubling a gas’s pressure will halve its volume if other variables are held constant. So the effect on volume of doubling pressure cancels out the effect of doubling Kelvin temperature, and the net result is that the gas’s volume will stay the same. After 1 half-life, a 100 g sample will have a mass of 50 g. After 2

36. A half-lives, it will have a mass of 25 g, which is the expiration of 2 half-lives. If 2 half-lives = 12.4 hours, then 1 half-life = 6.2 hours. 37. C 38. A This is alpha decay, in which a radioactive atom loses 2 protons 39. E and 2 neutrons. The loss of 2 protons means the atomic number decreases by 2, which means it was 87 before it decayed. The element with an atomic number of 87 is Fr (francium). The loss of 2 neutrons together with the loss of 2 protons means that the mass number has decreased by (2) + (2) = 4. The mass number is now 216, which means that it was 220. Notice that (C) has the right numbers, but not the right element. Remember that the atomic number uniquely identifies an element, so an atomic number of 87 must be francium, no matter what the mass number is. This also means that an element can have several different mass numbers, collectively called isotopes. The empirical formula tells you that the ratio of carbon to hydrogen is 1:2, so look for an answer that reflects the same ratio. Only (B), (C), and (E) do that, so eliminate (A) and (D). Now, among (B), (C), and (E), look for the one whose molecular weight is 56 amu. Look at the periodic table. Every carbon atom has an atomic weight of 12 amu, and every hydrogen atom has an atomic weight of 1 amu. Rather than pursue algebra, just try the three choices. Don’t fall into the temptation trap and pick (B). Temperature differences can indicate the direction of heat flow. However, temperature is not a direct measure of heat energy. Instead, associate heat energy with enthalpy. On this test, the easiest way to balance an equation is by plugging in the answers. The test writer tells you that the coefficient in front of H2O is 6, so put that in there. Now, see which of the answer choices, if placed in front of NF3, would result in a balanced equation. If, for instance, you try (B), you’d have 1 mole of N on the left, which would give 3 moles of F on the left. In order to have 3 moles of F on the right, you’d have to put a 3 in front of the HF

40. B on the right. That means you’d have 3 moles of H on the right and 41. C 12 moles of H on the left. This is way out of balance. 42. C Suppose you try option (D) and put a 3 in front of the NF3 on the 43. E left. That gives 9 moles of F on the left, which means you’d have to put a 9 in front of the HF on the right. That in turn would provide 9 moles of H on the right when you have 12 moles of H on the left. Once again, this is out of balance. Now, try (E). Put a 4 in front of NF3 on the left, which gives 12 F on the left and means that you must put a 12 in front of the HF on the right. That gives 12 H on the right, balanced by 12 H on the left. The coefficients for both NO and NO2 would be 2, which would balance both the nitrogens (N) and the oxygens (O). Make sure you can distinguish between an acid and a base. A base can donate an unshared electron pair, according to the Lewis definition, so (B) is correct. Choices (A), (D), and (E) are characteristics of acids. And what about (C)? Not all bases contain the OH– ion in their structure—NH3 is an example. The substance described has some metallic characteristics (shiny and high melting point) and some nonmetallic ones (brittle and poor electrical conductivity). It sounds like something that’s between a metal and a nonmetal such as a metalloid or semimetal. This is a general description of the metalloid silicon. The freer molecules are to move around, the greater the entropy of the state. Equation I goes from a gas (very high entropy) to a liquid (more ordered, less entropy). That’s a decrease in entropy. Equation II also involves an entropy decrease. Here, ions go from being able to move throughout a solution to being restricted in the solid state. Equation III shows an entropy increase because it is increasing the moles of gas (2 on left and 3 on right). I and II illustrate an entropy decrease, and the correct answer is (C). Choices (A) and (D) will not change water’s boiling point.

44. D Remember that when a solute is dissolved in solution, the solution’s boiling point will be raised (and its freezing point 45. C lowered). Only (E) involves dissolving a solute into water. Sugar 46. E water boils at a higher temperature than pure water under identical 47. D conditions. Why doesn’t the addition of gasoline into the water have the same effect? Gasoline molecules are nonpolar and will not dissolve in water. As for (C), increasing the altitude of the water will decrease its boiling point. The question concerns periodic table trends and, in particular, atomic radius. As you move from left to right across a period, atomic radius decreases. So, within a period, the higher the atomic number, the smaller the atomic radius. This question is a perfect example of the camouflage trap. You might be thinking higher atomic number, and the correct answer is phrased as “greater positive charge in its nucleus.” But keep the blinders off your brain. You know the answer—just remember that there’s more than one way of expressing it. Aluminum is a metal, and oxygen is a nonmetal. They react to form an ionic compound. Aluminum (in the 3A group) forms a +3 ion. Oxygen (in the 6A group) forms a –2 ion. They will produce aluminum oxide, Al2O3. When the equation is balanced, you’ll get 4Al(s) + 3O2(g) → 2Al2O3(s). Many colored compounds contain a transition metal (an element from the d region of the periodic table). Choices (A), (B), and (C) are ionic solids that possess an active metal (an element from the s region). These compounds appear white (for instance, NaCl or table salt). Choice (D) is rust, which is not blue. CuSO4 contains the transition metal copper (Cu), and its crystals are bright blue. Remember that the atomic weight of an element is the weighted average of all the different isotopes an element exists in. If 50 percent of element X had a mass of 210 amu and 50 percent had a mass of 214 amu, the weighted average would be 212 amu. Notice

48. B that you are told element X exists as 214X more than half of the 49. D time. So the answer must exceed 212. However, since element X 50. D also exists in an isotope with a mass less than 214, expect that its atomic weight is less than 214 amu. Only (D) has a mass greater 51. A than 212 amu and less than 214 amu. The container holds a total of 6 moles of gas. Oxygen (O2) constitutes one-third of that content. If you know how to work partial pressure problems, you know that oxygen’s contribution to the total 760 torr of pressure is one-third: 760 torr/3 = approximately 253 torr. Since you’re considering the dissolution of an ionic solid into water, you need to consider the solubility product expression. The solubility product constant, Ksp, will equal the product of aqueous ion concentrations raised to their coefficients. (Remember that solids are not expressed.) This is the relationship expressed in (D). The normal electron configuration for magnesium would be 1s22s22p63s2. Since the question is talking about an “excited” magnesium atom in which one electron has been pushed up into a higher energy state, look for a configuration that shows one electron in a higher state than it should be. The total number of electrons should still be equal to magnesium’s atomic number, but the location of one electron should be “elevated.” Choice (D) is just what you’re looking for. Total number of electrons? Twelve, just as it should be. But look at the last entry. Instead of 3s2, it has 3s13p1. The last electron has been elevated to the 3p subshell. To form an electrolytic solution, the solute must dissociate into ions. Adding HCl(g) to water will produce hydrochloric acid, which ionizes into H+ and Cl– ions. Choices (C), (D), and (E) are all ionic solids, which will break into mobile ions upon dissolution in water. When N2(g) is dissolved into water, no ions are produced, and the resultant solution is nonelectrolytic.

52. C The double arrow indicates that the reaction is reversible. NH4+ is 53. A a reactant of the reverse reaction; if NH4+ donates a proton to 54. B HCO3–, NH3 and H2CO3 are formed. Since NH4+ donates an H+ ion (or proton) to another substance, it acts as an acid according to 55. E the Brønsted-Lowry definition. Add the superscripts to get the total number of electrons in the species: 2 + 2 + 6 + 2 + 6 = 18. Which of the choices also has 18 electrons? A quick check of the periodic table shows that a sulfur atom has 16 electrons. Adding two more electrons gives the S2– ion a total of 18 electrons. For a substance to be amphoteric, it must be able to donate and receive an H+ ion. Eliminate (A) and (E)—these species don’t have an H+ ion to donate. Choice (C), KOH, is a strong base. You wouldn’t expect it to ever act as an acid. Likewise, HNO3 ((D)) is a strong acid that you would not expect to behave as a base. That leaves HSO4–. Notice that it can act as an acid and become a sulfate ion, SO42–, or act as a base and become sulfuric acid, H2SO4. The ideal gas equation is PV = nRT. When the amount of gas does not change, n becomes a constant, like R. A little algebra gives = nR. Since is equal to a constant, it will not change with time. Consider that you are dealing with two points in time. At first, the gas has a volume of 10 liters at 20°C or (293 K) and 750 mmHg. So here, . Later, the gas is under STP conditions, so T = 273 K and P = 760 mmHg. Now, . If you set initial = final , you get . Rearranging and

applying some algebra gives V = 10 × L. This is (E). 56. A Use the phase diagram. At 0.5 atm and 200 K, substance Z is a gas. 57. C If you maintain this temperature and increase the pressure, you can 58. B draw a vertical line from the point (0.5 atm, 200 K). Eventually that vertical line will cross into the liquid region. This means that, under steadily increasing pressure, substance Z (starting at 0.5 atm and 200 K) will condense. Condensation is the phase change from gas to liquid. The normal boiling point is the temperature at which the phase change from liquid to gas occurs, at a pressure of 1 atm. If you extend a horizontal line from the 1.0 atm mark on the “pressure” axis and see where it intersects the liquid-gas boundary, you’ll get the normal boiling point. Doing so on this phase diagram shows a normal boiling point of about 300 K. Phosphorus is the central atom in PCl3. A phosphorus atom needs 3 electrons to complete its valence shell. It gets 3 electrons by forming covalent bonds with 3 chlorine atoms. The PCl3 molecule has the following structure: 59. A Of the four electron pair sites around phosphorus, one is a lone pair. This gives the PCl3 molecule a trigonal pyramidal shape. First, consider the neutralization that occurs between HBr and Ba(OH)2. 2HBr + Ba(OH)2 → BaBr2 + 2H2O Notice that for every 2 moles of HBr, only 1 mole of Ba(OH)2 is needed for neutralization. You have 0.1 liters (or 100 milliliters) of 0.2 M HBr. This means you have 0.1 liters × 0.2 mole/liter, or 0.02

60. A mole of HBr. You need 0.01 mole of Ba(OH)2 to neutralize 0.02 mole of HBr. Twenty-five milliliters of 0.4 M Ba(OH)2(aq) has 61. E 0.025 liters × 0.4 mole/liter, or 0.01 mole of Ba(OH)2. 62. A A small Ka indicates a weak acid. That means statement III is false, 63. D and therefore you can eliminate (D) and (E). It also means that 64. E most HCN remains as intact molecules, as opposed to H+ and CN– ions, so statement II is false. POE tells you that statement I must be true. And it is: HCN ⇋ H+ + CN–. Notice that the molar ratio of H+ to CN– is 1:1. Ionization energies get very large once you try to remove core electrons, which are attracted more strongly to the nucleus than valence electrons. So an atom with a very high second ionization energy would be expected to have just 1 valence electron: The second electron to be removed from such an atom would have to be a core electron. Among the choices, only sodium atoms have a single valence electron. When the reaction is flipped, the sign flips and become negative. Also, if all of the coefficients on the substances in the reaction are divided by two, then so is the value. Remember that many metals react with acids to produce hydrogen gas. Your first step should be to write out the reaction. Here it is: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). Notice that the products of this reaction are Zn2+(aq), Cl–(aq) (from ZnCl2), and H2(g). This is a Le Châtelier’s principle question. How can you produce crowding on the left side of the equation and drive equilibrium to the right (increasing the SO2 concentration)? Don’t be fooled by (C): Catalysts do not affect equilibrium, but changing concentrations does influence equilibrium. Increasing the concentration of O2 will produce crowding on the left side and lead to an increase in the concentration of SO2.

65. A Decreasing the volume of the system will increase the concentration of reactants. Why? Because the same number of 66. B molecules now exists in a smaller space, and this increases the 67. E ratio of molecules per volume. How will this affect the reaction 68. A rate? It will increase reaction rate because decreasing the volume makes it more likely that molecules will collide. What about equilibrium? Reducing the volume will force the equilibrium to shift in the direction that produces fewer moles of gas. This means the equilibrium concentration of reactants will decrease because equilibrium will shift to the right (4 moles of gas on the right versus 5 on the left). Only a temperature change will affect the value of Keq, so this will stay the same. Among the three, only item I will increase, so the answer is (A). When quantities are given for more than one reactant, you must see which is limiting. Fifty-six grams of CO (molecular weight = 28 amu) are 2 moles of CO. Since the stoichiometric ratio of Fe2O3 to CO is 1:3 (based on coefficients from the balanced equation), CO is limiting. (You would need more than 9 moles of CO based on the 1:3 ratio for Fe2O3 to be limiting in this case.) The ratio of CO to Fe is 3:2. So 2 moles of CO will produce about 1.3 moles of Fe. You could quickly estimate this as between 1 and 2 to save time and still get the answer. Consider what gases are being collected in the beaker. Oxygen gas is flowing in from the reaction. Water vapor, H2O(g), is also entering the beaker from the evaporation of water. The pressure exerted by H2O(g) is equal to the vapor pressure of water. The total gas pressure in the beaker is, therefore, the sum of the pressure of oxygen gas collected and vapor pressure of water (at 25°C in this particular problem). If you draw the structure of acetylene, C2H2, you’ll see that the carbon atoms must share three pairs of electrons to achieve stable octets. H – C ≡ C – H

69. D So C2H2 has a triple bond. None of the other molecules contains a 70. D triple bond. Since this is the strongest type of carbon-to-carbon bond, (A) is correct. Equilibrium is attained when the concentrations of all species become constant. The concentrations of N2O4 and NO2 stay the same from the 20-minute mark to the 25-minute mark. This means equilibrium was achieved before the 20-minute mark. Since the concentrations of N2O4 and NO2 are different from the 15-minute mark to the 20-minute mark, equilibrium was not achieved at exactly 15 minutes from the start of the reaction; equilibrium was attained between 15 and 20 minutes after the start of the reaction. An emission spectrum is created when electrons in an excited state fall back to their ground state. For the electrons to become excited, an outside energy source is required to overcome their attraction to the nucleus.

HOW TO SCORE PRACTICE TEST 1 When you take the real exam, the proctors will collect your test booklet and bubble sheet and send your answer sheet to a processing center, where a computer looks at the pattern of filled-in ovals on your answer sheet and gives you a score. We couldn’t include even a small computer with this book, so we are providing this more primitive way of scoring your exam. Determining Your Score STEP 1 Using the answer key, determine how many questions you got right and how many you got wrong on the test. Remember: Questions that you do not answer don’t count as either right or wrong answers. STEP 2 List the number of right answers here. (A) ________ STEP 3 List the number of wrong answers here. Now divide that number by 4. (Use a calculator if you’re feeling particularly lazy.) (B) ________ ÷ 4 = (C) _______ STEP 4 Subtract the number of wrong answers divided by 4 from the number of correct answers. Round this score to the nearest whole number. This is your raw score. (A) ________ – (C) ________ = _________ STEP 5 To determine your real score, take the number from Step 4 above, and look it up

in the left column of the Score Conversion Table on the next page; the corresponding score on the right is your score on the exam.

PRACTICE TEST 1 SCORE CONVERSION TABLE



Part III Subject Review 4 Atoms: The Building Blocks of Matter 5 Chemical Reactions and Stoichiometry 6 Electron Configurations and Radioactivity 7 The Periodic Table and Bonding 8 Phases: Gases, Liquids, and Solids 9 Solutions 10 Kinetics and Equilibrium 11 Acids and Bases 12 Oxidation and Reduction 13 Organic Chemistry and Environmental Chemistry 14 Laboratory

Chapter 4 Atoms: The Building Blocks of Matter Atoms are the building blocks of matter. Understanding the structure of atoms and how atoms of different elements and isotopes differ from one another is critical to success on the SAT Subject Test in Chemistry. Atoms and elements are the focus of this chapter.

ATOMS AND ELEMENTS An element is any substance that cannot be broken down into a simpler substance by a chemical reaction. Now, what exactly is an atom? Suppose you have a spoonful of some element— carbon, for instance. The smallest, tiniest, teeniest “piece” of carbon in the spoonful is one atom of carbon. Technically, an atom is the smallest particle of an element that still retains the chemical properties of that element. Small Particle, Big History The concept of an atom has been kicking around since ancient Greece, but it wasn’t until the early 1800s that the English scientist John Dalton supported these theories with measurements. His proposal was that all elements were made up of atoms and, furthermore, that all atoms of a given element were identical to one another. He also stated that atoms of different elements therefore had different masses and that a chemical reaction would involve the rearrangement of atoms. How an Atom Is Made: Protons, Neutrons, and Electrons At the center of every atom is a nucleus. What makes up the nucleus of an atom? Two things: protons and neutrons. Protons have a charge of +1, and neutrons have no charge at all—they are neutral. Because protons and neutrons comprise the nucleus of an atom, they are sometimes referred to as nucleons. What’s outside the nucleus? Electrons; electrons have a charge of –1. In an electrically neutral atom, the number of electrons is equal to the number of protons; the charges inside and outside the nucleus are balanced.

Sometimes it happens that an atom loses or gains one or more electrons. When that happens, the number of electrons outside the nucleus is not equal to the number of protons inside. The atom isn’t electrically neutral anymore and has become an ion. If the atom loses one or more electrons, it has fewer negative charges than positive charges, so it is a positively charged ion, or cation. If the atom gains one or more electrons, it has more negative charges than positive charges. This is a negatively charged ion, or anion. Seeing Is Believing One of the major advances in determining atomic structure came with the creation of a cathode ray tube (CRT). A CRT is essentially an electron gun—a device capable of producing electrons and shooting them through a vacuum, along with a screen that is able to detect the electrons. Scientists originally determined the rays created by a CRT were negatively charged, as they were attracted to positive objects and repelled from negative ones. The first scientist to determine the rays were actually made up of particles was J.J. Thomson, who also determined the relative mass and charge of the particles. These particles later came to be known as electrons. The most famous experiment that helped shape modern atomic theory was

conducted by Ernest Rutherford. Rutherford shot a beam of helium atoms at a sheet of gold foil. The vast majority of atoms passed through undeflected, which was expected—atoms were considered to be mostly empty space, a theory that is still held to be true today. However, a very small percentage (<1%) of the atoms were deflected by the foil. Rutherford theorized that this was due to the presence of a highly dense nucleus located in the center of the atom and that had sufficient mass to deflect the helium atoms. The theory of the atom continues to evolve over time to expand beyond an understanding of the basic subatomic particles, but that will be addressed further in Chapter 6. THE PERIODIC TABLE When you sit down to take the SAT Subject Test in Chemistry, you will be provided with a periodic table. On the periodic table, the vertical columns are called groups, and the horizontal rows are called periods. The symbols (such as H, Li, Be, etc.) represent elements. PERIODIC TABLE OF THE ELEMENTS

For the SAT Subject Test in Chemistry, you have to know certain things about the elements and other information you see on the periodic table. Atomic Number Look at any element on the periodic table. Above every element’s symbol is a whole number that represents the element’s atomic number. The atomic number of an element tells you the number of protons in the nucleus of an atom of that element. The number of protons in the nucleus of an atom gives that atom its identity. Oxygen, for instance, has the atomic number 8; an atom of oxygen has 8 protons in its nucleus. If you take a proton away from an oxygen atom, it would have only 7 protons in its nucleus, and it wouldn’t be oxygen anymore; it would be a nitrogen atom. Any atom with 7 protons in its nucleus is nitrogen.

What if an oxygen atom loses an electron but not a proton? Well, as long as the atom has 8 protons in it, it’s still oxygen. If it has only 7 electrons, then it’s a positively charged oxygen ion, and if it has 9 electrons, then it’s a negatively charged oxygen ion. But as long as the atomic number—the number of protons in the nucleus—doesn’t change, the element doesn’t change either: It’s still oxygen. Atomic number: number of protons in nucleus; identity of element Group Number The group number refers to the number above each column of the periodic table, such as 1A, 2, 4, etc. Elements in the same group have the same number of electrons in their outer shells and share many properties. Carbon (C), Silicon (Si), and Germanium (Ge), for example, are all in the group 4A and have similar chemical properties. Mass Number, Isotopes, and Atomic Weight We said before that the nucleus of an atom is made up of two types of subatomic particles: protons and neutrons. Protons have a positive charge, and neutrons are neutral. Another important point is that both protons and neutrons have mass, while electrons have practically no mass. Protons and neutrons each have mass of roughly 1 atomic mass unit (amu), and an atom’s mass number is equal to the sum of its protons and its neutrons. Now, we’ve said that in an electrically neutral atom, the number of protons is equal to the number of electrons. But this says nothing about neutrons. Most carbon atoms, for instance, have 6 neutrons in their nuclei. A few have 8. All have 6 protons in their nuclei, so they all have the atomic number 6, and they’re all carbon. But they can differ in the number of neutrons. If two atoms of the same element differ in the number of neutrons in their nuclei,

they are said to be isotopes; a carbon atom that has 6 neutrons and a carbon atom that has 8 neutrons are isotopes. Since an atom’s mass number is equal to the sum of its protons and its neutrons, two different isotopes of the same element will have different mass numbers. For instance, the carbon atom that has 6 neutrons in its nucleus has a mass number of 6 protons + 6 neutrons = 12 amu. And, because its mass number is 12, we call it carbon-12. The carbon atom that has 8 neutrons in its nucleus has a mass number of 6 protons + 8 neutrons = 14 amu. And, because its mass number is 14, we call it carbon-14. Let’s Talk Numbers By the way, mass number doesn’t appear on the periodic table. Why not? Because for any element, there’s no such thing as one mass number. As we’ve just learned, different isotopes of the same element have different mass numbers. However, chemists have figured out, roughly, the degree to which each isotope of each element tends to occur on Earth. So, for each element, they’ve figured out an average mass number, which for each element represents the average of the mass numbers of all isotopes as they occur on Earth. This average number is called the atomic weight of the element. The atomic weight of each element appears on the periodic table, just below the element’s symbol. When we want to know the mass of an atom of a particular element, for practical purposes, we use the atomic weight that appears on the periodic table.

Remember the following: Neutrons: • along with protons, make up the nucleus of an atom • electrical neutrality • isotopes—two atoms of the same element are isotopes if they differ in their number of neutrons Mass • number of protons plus number of neutrons; number: • isotopes—two isotopes of an element have different mass numbers because they differ in the number of neutrons Atomic • average of mass numbers for all isotopes of an element as weight: they occur on Earth Now try the following questions. The answers can be found in Part IV.

DRILL 1 Question Type A Questions 4-8 refer to the following. (A) Atom (B) Ion (C) Neutron (D) Proton (E) Electron 4. The smallest representative particle of helium 5. Its loss or gain creates a positively or negatively charged ion, respectively 6. Particle responsible for positive nuclear charge 7. Isotopes of uranium always differ in their number of this particle 8. Their number in the nucleus determines an element’s atomic number Question Type B II I

105. The periodic table does BECAUSE a mass number can be assigned not report mass to one isotope of an element but numbers not to an element in general. 106. Addition of an electron BECAUSE every electron carries a negative to an atom creates a charge. positively charged ion Question Type C 27. Two different sodium atoms or ions may differ in all of the following ways EXCEPT (A) the number of electrons outside their nuclei (B) the overall charge they carry (C) their mass numbers (D) the number of neutrons in their nuclei (E) the number of protons in their nuclei 28. Two isotopes of the same element will always differ in (A) mass number but never in atomic number (B) atomic number but never in mass number (C) charge outside but never inside their nuclei (D) nuclear charge but never in overall charge (E) the number of electrons outside their nuclei but never in the

number of neutrons inside their nuclei

Summary ○ An atom consists of neutrons (charge = 0, mass = 1 amu) and protons (charge = +1, mass = 1 amu) in a nucleus orbited by electrons (charge = –1, mass = 0 amu). ○ Atomic number is the number of protons in an atom of an element, and it defines the element. ○ Mass number is the number of protons plus the number of neutrons in an atom. ○ Isotopes are atoms of the same element that contain different numbers of neutrons. ○ Atomic weight is the average of the masses of the different isotopes based on the frequency with which they occur in nature.

Chapter 5 Chemical Reactions and Stoichiometry Chemistry is a science largely concerned with reactions, or how molecules of one or more substances are converted to molecules of a different substance(s). This chapter discusses formula (or molecular) weight, empirical formulas, and percent composition; moles and molar mass; balancing chemical reactions and stoichiometry; and thermodynamics, which is the role of energy in a reaction: how much energy is given off or absorbed by a reaction and what this can tell us about the reaction.

THE WORLD OF MOLECULES As we mentioned in Chapter 4, an atom is the smallest particle of an element. Atoms of one element often attract each other or atoms of a different element. If this attraction is strong enough, a chemical bond can result. Chemical bonds join two or more atoms into units called molecules. (We’ll talk more about bonding in Chapter 7.) Everybody knows the formula of water—it’s H2O. This formula tells us that, in a molecule of water, 2 hydrogen atoms and 1 oxygen atom are bonded together into a unit. An individual water molecule is the smallest unit of water that exists. Diatomic Molecules When a molecule consists of just two atoms (whether they are of identical or different elements), it’s called a diatomic molecule. Some elements exist as diatomic molecules at room temperature and atmospheric pressure (1 atm). For example, oxygen in air exists in the form of O2. There are seven important elements that exist as diatomic molecules: oxygen (O2), iodine (I2), hydrogen (H2), nitrogen (N2), chlorine (Cl2), fluorine (F2), and bromine (Br2). Elements as Diatomic Molecules One good way to remember all of the diatomic molecules is the following mnemonic: Oh I Have Nice Closets For Brooms O2, I2, H2, N2, Cl2, F2, Br2 Formula Weights, Empirical Formulas, and Percent Composition

For the SAT Subject Test in Chemistry, you’ll need to know some simple atom- molecule math. 1. Formula Weight: For any molecule, we calculate the formula weight by adding up the atomic weights of all the atoms in the molecule. It’s easy. Take hydrogen peroxide (H2O2) as an example. • The molecule has 2 hydrogen atoms and 2 oxygen atoms. Let’s start with hydrogen. • The atomic weight of hydrogen is 1 amu. Since there are 2 atoms, we have a total of 2 × 1, or 2 amu. • Now let’s talk about the oxygen in the molecule. The atomic weight of oxygen is 16, and since there are 2 atoms, the total is 16 × 2, or 32 amu. • The total is 2 + 32 = 34. The formula weight for H2O2 is 34 amu. Here’s another example. Let’s figure out the weight for sulfuric acid: H2SO4. • The molecule has 2 hydrogen atoms, 1 atom of sulfur, and 4 oxygen atoms. Let’s start with hydrogen. • The atomic weight of hydrogen is 1 amu. We have 2 hydrogens, so there are a total of 2 amu. • The atomic weight of sulfur is 32 amu. So we have 32 amu of sulfur. • Now let’s talk about the oxygen. We know the atomic weight of oxygen is 16 and there are 4 atoms. So we have a total of 16 × 4, or 64 amu. • 2 + 32 + 64 = 98. The formula weight for H2SO4 is 98 amu. 2. Empirical Formula: An empirical formula shows the ratio of atoms within a molecule. To find an empirical formula from a molecular formula, first find the largest whole number by which all of the subscripts in the molecular formula are divisible. Then divide each subscript by that number. Let’s look at a few examples. • We talked about the molecular formula of hydrogen peroxide, which is H2O2. The largest number that can divide into 2 is 2. If we factor a 2 out of both subscripts, we’re left with HO. So the empirical formula for hydrogen peroxide is HO.

• Ethane has the molecular formula C2H6. Let’s find the empirical formula. What’s the largest number that goes evenly into 2 and 6? It’s 2. If we divide both subscripts by 2, we get an empirical formula of CH3. • How about water? The largest whole number that will divide evenly into 2 and 1 is 1. So the empirical formula for water is the same as its molecular formula: H2O. 3. Percent Composition: This refers to the percent of a substance contained within a molecule. The test writers may ask you to determine this percent. Here is one example of a question you might see on the test. Percent by Mass Whenever a question asks about the percent composition, the implication is the percent by mass of the substance. What is the percent of oxygen by mass in hydrogen peroxide, H2O2? In order to determine the percent composition of oxygen, you need to find the mass in amu of all of the oxygen atoms in the molecule and compare it with the total formula weight. As shown earlier, each of the two oxygen atoms in H2O2 weighs 16 amu. This means that the mass due to oxygen is 2(16) or 32 amu. The formula weight of H2O2 is 34 amu. So the percent of oxygen by mass in H2O2 is × 100%, or roughly 94%. You can then calculate the percent of hydrogen by mass in H2O2, which is 100% – 94%, or about 6%. THE MOLE What’s a mole? It’s a number, like a dozen or a gross. Dozen, as you know,

means 12. Gross means 144. Mole means 6.02 × 1023. The number 6.02 × 1023 is known as Avogadro’s number. We said earlier that atomic mass is measured in atomic mass units (amu); 1 amu is 1/12 the mass of a carbon-12 atom, or a relative mass of 1 gram (g). Each 1 g of a substance has 6.02 × 1023 amu. Think about what that means. If, for any element, we take a sample whose mass in grams is numerically equal to its atomic weight in amu, the sample has 1 mole of atoms in it. If we take a substance whose mass in grams is numerically equal to twice its atomic weight in amu, we have 2 moles of atoms. If we take a substance whose mass in grams is numerically equal to three times its atomic weight in amu, we have 3 moles of atoms. It’s as simple as that. For example, helium’s atomic weight is 4 amu, so 4 g of helium contain 1 mole (6.02 × 1023) of helium atoms, and 8 g of helium contain 2 moles of helium atoms. Carbon’s atomic weight is 12 amu, so in 12 g of carbon there is 1 mole (6.02 × 1023) of carbon atoms, and in 36 g of carbon there are 3 moles (18.06 × 1023) of carbon atoms. Got it? How many moles of oxygen molecules are in 64 g of oxygen gas? Remember that oxygen is diatomic. Each O2 molecule has a mass of roughly 2(16), or 32 amu, so 1 mole of O2 molecules would have a mass of 32 g. Thus, in 64 g of oxygen gas, there are 2 moles of oxygen molecules. Now how many moles of oxygen atoms would be present in this 64 g sample? Each O2 molecule is made up of 2 oxygen atoms, so if the sample contains 2 moles of oxygen molecules, it contains 4 moles of oxygen atoms. Note that the lower the molar mass is for a given element, the more moles there will be present in 1.0 g of it. Hydrogen has a molar mass of 1.0 g/mol, and helium has a molar mass of 4.0 g/mol. 1.0 g of hydrogen atoms would thus represent one mole of hydrogen, but 1.0 g of helium atoms is only 0.25 mol of helium! This is important—moles represents the amount of substance that reacts (or is produced) during a chemical reaction, and all calculations in chemistry revolve around moles!

Converting Mass Composition to Empirical Formula Now that you know what an empirical formula is, we’ll tell you how to figure out an empirical formula from the percent composition of a molecule. For example, the test writers might tell you that some unknown substance is made up of approximately 75% mercury and 25% chlorine, and they might ask you to take this percent composition and figure out the substance’s empirical formula. Here’s how you do it. 1. Imagine, first, that you have 100 g of the substance. 2. If you have 100 g of the substance, and it’s 75% mercury by mass, then you’ve got 75 g of mercury, right? Since the atomic weight of mercury is about 200 amu (which means that 1 mole of mercury atoms weighs 200 g), you’ve got mole = 0.375 moles of mercury atoms in a 100 g sample. 3. If you have 100 g of the substance, and it’s 25% chlorine by mass, then you’ve got 25 g of chlorine. Since the atomic weight of chlorine is about 35 amu (which means that 1 mole of chlorine atoms weighs 35 g), you’ve got = 0.700 moles of chlorine atoms in the 100 g sample. 4. If you’ve got 0.375 moles of mercury atoms and 0.700 moles of chlorine atoms, then the ratio of chlorine to mercury atoms is (which is close to 2:1), which means the empirical formula is HgCl2. Finding Moles When finding moles of a diatomic element such as chlorine in a compound, use its atomic weight, not its formula weight in the calculation.

Now review everything we’ve told you about molecules and moles, and answer the following questions. The answers can be found in Part IV.

DRILL 1 Question Type A Questions 1–4 refer to the following. (A) N2O (B) C6H12O6 (C) SO3 (D) NO (E) N2O5 1. Is a diatomic molecule 2. Has a formula weight of approximately 108 amu 3. Has an empirical formula that is different from its molecular formula 4. Composition is approximately 60% oxygen by mass Question Type B I II 101. Chlorine is an element BECAUSE chlorine exists as unbonded atoms at room temperature and

atmospheric pressure. 102. One mole of HBr has BECAUSE the mass of a molecule of HBr is greater mass than one greater than the mass of a mole of NO2 molecule of NO2. Question Type C 24. The formula for calcium nitrate is Ca(NO3)2. What is its approximate formula weight? (A) 64 amu (B) 164 amu (C) 240 amu (D) 310 amu (E) 380 amu 25. An unknown substance is found to have a composition of 9% magnesium and 91% iodine by weight. The empirical formula for the substance is (A) MgI (B) Mg2I 2 (C) Mg2I (D) MgI2

(E) Mg3I2

CHEMICAL REACTIONS—HOW MOLECULES ARE FORMED, BROKEN DOWN, AND REFORMED In the course of a chemical reaction, the bonds that hold together the atoms that make up the reactants break. The free atoms then form new bonds with one another to form new molecules—the products of the reaction. Take a look at the chemical reaction below: C3H8(g) + 5O2(g) → 4H2O(l) + 3CO2(g) What does this equation tell us? Well, two things. On a molecular level, this equation says that 1 molecule of propane (C3H8) and 5 molecules of oxygen react to form 4 molecules of water and 3 molecules of carbon dioxide. It also tells us that 1 mole of C3H8 reacts with 5 moles of O2 to form 4 moles of H2O and 3 moles of CO2. This equation also indicates the state of each reactant and product: (s) means solid, (l) means liquid, and (g) means gas. Note that when a substance is dissolved in water to create a solution, it is in the aqueous phase, represented with (aq). This is not the same thing as the liquid phase. If you were to take table salt (NaCl (s)) and heat it until it melted, it would become NaCl (l). However, if you took that table salt and dissolved it into a glass of water, it would create NaCl (aq). Much of chemistry is solution-based, and compounds will frequently be dissolved in water to form aqueous solutions before any reactions take place. Chemical Equations Must Be Balanced Look again at the chemical equation we just presented:

C3H8(g) + 5O2(g) → 4H2O(l) + 3CO2(g) This is a balanced equation. How can you tell? For each element on the left side of the equation, multiply the molecular coefficient by the element’s subscript. (Any number that doesn’t appear is assumed to be 1.) For oxygen, in 5O2, there are 5 × 2 = 10 oxygen atoms. Now do the same for the right side of the equation. In 4H2O, there are 4 × 1 = 4 oxygen atoms. In 3CO2, there are 3 × 2 = 6 oxygen atoms. So there are 4 + 6 = 10 oxygen atoms on the right side, and since there are also 10 oxygen atoms on the left side, oxygen is balanced. Now check to see that carbon and hydrogen are also balanced. They are. There are 3 carbons on the left and 3 carbons on the right. There are 8 hydrogens on the left and 8 hydrogens on the right. For each element in a balanced equation, the total number of atoms on the left must equal the total number of atoms on the right. Balancing Equation Steps: 1. First, start with the smallest answer choice, and plug it in front of what you want to solve for. 2. Next, turn to the other side of the equation and fill in coefficients based on what you started with. 3. If you can’t find whole number coefficients so all the atoms on either side of the equation add up, turn to the next smallest answer choice and repeat. 4. If you can balance the equation, check to make sure there is no common factor for all of the coefficients. For example, 4, 6, and 10 have a common factor of 2. If there is a common factor, divide the coefficients, and you have your answer. If there isn’t, you already have your answer. On the SAT Subject Test in Chemistry, you may see up to five questions that will show you unbalanced equations and ask you to balance them. Here’s what those will look like. 29….C2H4(g) +…O2(g) →…CO2(g) +…H2O(l)

When the equation above is balanced and all coefficients are reduced to lowest whole-number terms, which of the following would be the coefficient for CO2 ? (A) 1 (B) 2 (C) 4 (D) 5 (E) 6 Fortunately, these questions are easy to answer if you use the “plug-in” balancing strategy. Start with (A), and put the number 1 in front of CO2. (Don’t be afraid to write in your test booklet as much as you want. Your test booklet belongs to you, and nobody cares what you write in it; only your answer sheet is scored.) By adding a 1 in front of CO2, we end up with 1 carbon on the right. Yet, we have at least 2 carbons on the left, so we know that 1 is not the answer. Let’s try (B). Put a 2 in front of CO2, and see what happens. We have 2 carbons on the right and 2 on the left. Good. We have 4 hydrogens on the left, so let’s try putting a 2 in front of the H2O. That gives us a total of 6 oxygens on the right and 2 on the left. Let’s put a 3 in front of O2 on the left, so we have a total of 6 oxygens on the left. Now we have 2 carbons on the right and the left, 4 hydrogens on the right and the left, and 6 oxygens on the right and the left. The equation is balanced, and (B) is correct. So, every time a question asks you to balance an equation, use the plugging-in strategy. It can’t fail. STOICHIOMETRY Sometimes SAT Chemistry questions will ask you to determine how much product is formed or reactant is consumed in the course of a chemical reaction. These are stoichiometry questions. When you begin a stoichiometry problem, always remember to work from a balanced equation. The coefficients in front of each species indicate the molar ratio between the species. Consider the reaction between ammonia gas and oxygen, which yields nitrogen monoxide and water.

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) This equation tells us that • For every 4 moles of ammonia consumed, 5 moles of oxygen are also consumed. • For every 5 moles of oxygen consumed, 6 moles of water are produced. • For every 4 moles of ammonia consumed, 4 moles of nitrogen monoxide are produced. In other words, the molar ratio of ammonia consumption to nitrogen monoxide production is 1:1. How do you put these molar ratios to use? Take a look: If 2 moles of ammonia are consumed, how many moles of water are produced? The best way to figure this problem out is to use unit conversions. The number of moles in our conversion comes from the coefficients of the balanced equation. 2 moles NH3 × = 3 mol H2O We know the moles of NH3 cancel because they appear in both the numerator and denominator, leaving us with moles of H2O. It can get a bit more complicated; we can only use the coefficients in the balanced equation to determine the mole ratio, not the mass ratio. So, if we start with mass, we have to convert to moles first. After we get our moles of product, we can then convert those to the mass of product. Consider this: if 34 g of NH3 gas reacts, how many grams of NO(g) are created? This time we have three necessary conversions: Note that when converting from grams to moles, we use the molar mass of each compound, which is always equivalent to the mass of one mole of that

compound. The coefficients from the balanced equation only appear in the mole ratio step, and never in the moles to grams step. The last thing that you need to know about stoichiometry (for now!) is that if you start with the amount of both reactants, inevitably, one of them will run out first. Once one of the reactants is gone, the reaction will stop. The reactant that runs out first is called the limiting reactant, and the reactant which does not run out is called the excess reactant. Close, But NO Cigar It’s rare in a laboratory to get 100% of the expected product, as a variety of experimental conditions can cause a loss. The percent yield in a reaction describes the percentage of the product that was successfully created, which is written as . So, in this example, if only 20 g of NO were produced, the percent yield would be = 83%. You can mathematically determine which reactant is limiting by seeing which would produce LESS product. Whichever reactant is able to produce less product will always run out first and thus be limiting. When 3 mol of NH3 is mixed with 3 mol of O2, the limiting reactant can be determined as follows: In this case, the oxygen would be the limiting reactant (and 2.4 moles of NO would be produced). We could have gone to moles of H2O instead of NO if we wanted to; it doesn’t matter which product we choose as long as we’re consistent. It’s important to realize that it’s not always the reactant which is present in lower amounts that will limit. The limiting reactant depends on a number of variables, including the molar mass of each reactant and the mole

ratio between the reactants and the products. THERMODYNAMICS Entropy One of the most important things to remember about thermodynamics is that low-energy states are more stable than high-energy states. That’s such a fundamental principle that we’ll ask you to repeat it. Fill in the blank lines: _________energy states are more stable than_________energy states. Fundamentally, the universe prefers low-energy states, and also fundamentally, it tends toward disorder. When we talk about disorder, we use the term entropy, which is symbolized by S. Everything tends toward maximum entropy. When we talk about a chemical reaction and the difference between entropy of the products and entropy of the reactants, we use the symbol ∆S. If ∆S is negative, the reaction has lost entropy; the products are more “orderly” than the reactants. If ∆S is positive, the reaction has gained entropy; the products are less “orderly” than the reactants. The State of Things Entropy is a measure of randomness, or disorder. To get an idea of what this means, consider the two boxes below. Which one has greater entropy, or disorder? It’s the one on the right.

The universe, it turns out, likes disorder. Systems move toward increasing disorder. They also tend to move toward states of lower energy. For the SAT Subject Test in Chemistry, when you see the word entropy, think “disorder,” and realize that, because the universe is lazy, it tends toward maximum entropy. All things in the universe are more stable when they’re in (1) states of low energy, and (2) states of high entropy. Higher Entropy + Lower Energy → More Stability Lower Entropy + Higher Energy → Less Stability Enthalpy Because the universe tends toward lower energy, chemical reactions that release energy—reactions that set energy free—are favored in the universe. When we talk about the energy states of reactants or products, we use the term enthalpy, which is symbolized by H. High enthalpy means high-energy state, and low enthalpy means low-energy state. So, the universe likes reactions in which the enthalpy decreases—reactions in which ∆H (the change in enthalpy that occurs in the course of a reaction) is negative. These reactions are said to be exothermic, and they result in the release of energy in the form of heat. If, however, the enthalpy of the products is greater than the enthalpy of the reactants, then ∆H is positive and the reaction is said to be endothermic. Endothermic reactions require the input of energy in order to take place. Energetic Bonds When we say that the products are at a higher or lower energy than the reactants, we mean the bonds in the products are at a higher or lower energy. If energy is released in a reaction, that energy comes from the potential energy stored in the bonds of the reactants. If the bonds of the products have more potential energy than the bonds of the reactants, we

have to supply the difference, and energy is absorbed in the reaction. Exothermic reaction → energy is released → ΔH < 0 → enthalpy decreases Endothermic reaction → energy is absorbed → ΔH > 0 → enthalpy increases Enthalpy is often represented graphically by a diagram often called a reaction coordinate. The reaction coordinate for an exothermic reaction looks like this: And an endothermic reaction would look like this:

Note that all that matters here is the energy difference between the reactants and the products. The “hump” in between them is related to the reaction’s speed, something that we’ll take a look at in a later chapter. Heat of Formation Another term you should be familiar with for the test is heat of formation—that is, the amount of heat that’s released or absorbed when 1 mole of a compound is formed from its elements. When we talk about heat of formation, we use the same symbol we use for enthalpy change, but we put a subscript “f” on it: ∆Hf. Let’s consider the heat of formation of gaseous carbon dioxide, CO2. C(s) + O2(g) → CO2(g); ∆Hf = –393 kJ/mol The fact that the heat of formation is negative means that heat is released during this reaction; this is an exothermic reaction. When 1 mole of CO2 (g) is formed from its elements (C(s) and O2(g)), 393 kJ of energy are released.

C, Ni, Cl2, O2, H2, and N2—or all other elemental atoms or molecules—have a heat of formation of zero. Remember this for the exam. For all elements, the heat of formation is zero. For this test, you’ll also need to keep in mind that for any reaction, the heats of formation of all the products minus the heats of formation of all the reactants is equal to ∆Hf for the whole reaction. The test writers might show you a reaction and give you heats of formation for all of the reactants and products. Then they’ll ask you to figure out ∆Hf for the whole reaction. That’s simple to do; you just add up the heats of formation for all of the products and then all of the reactants, multiplying each by its coefficient from the balanced equation, and you’ve got ∆Hf for the reaction. Remember that ∆Hf (reaction) = ∆Hf (products) – ∆Hf (reactants) and that the heats of formation of all elements are zero. Look at this reaction: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Suppose you’re told that the heat of formation for • C6H12O6(s) is –1,273 kJ/mol • H2O(l) is –286 kJ/mol • CO2(g) is –393 kJ/mol (∆Hf for O2(g), of course, is 0.) ∆Hf for the whole reaction is equal to ∆Hf(products) – ∆Hf(reactants) So ∆Hf(products) = 6 mol(–393 kJ/mol) + 6 mol(–286 kJ/mol) = – 4,074 kJ

∆Hf(reactants) = –1,273 kJ + 0 kJ = –1,273 kJ So, ∆H for the whole reaction = (– 4,074) – (–1,273) = –2,801 kJ. ∆H for the whole reaction is negative, which means the reaction is exothermic. We can also combine stoichiometry with thermochemistry in order to figure out how much heat is given off or released when a certain amount of reactants are used up. Let’s use the reaction we just looked at to show you how. If 18.0 g of C6H12O6 were to be reacted, how much energy would be released? Note that the –2,801 kJ goes with 1 mole of C6H12O6 because that is the coefficient on the C6H12O6 in the balanced reaction. If you were to instead start with 32.0 g of oxygen, the math would look like this: Remember, without a calculator, any math you would be asked to do should be fairly straightforward, as it is above. Hess’s Law If you have to manipulate any reaction, then there are rules in place for how the enthalpy of that reaction changes as well. There’s three rules you should know: 1. If you multiply (or divide) all of the species in that reaction by a coefficient, the enthalpy change must be multiplied (or divided) by the same coefficient. 2. If you flip a reaction, you must also flip the sign on the enthalpy. 3. If a reaction is the sum of several component reactions, the enthalpy for that reaction will be the sum of the enthalpies of the various components.

These rules combined are often called Hess’s Law. What Hess’s Law essentially tells us is that ΔH for a reaction is independent of the number of steps or the pathway the reaction follows. Take a look at the following example. Given: C(s) + O2(g) → CO2(g) ΔH = x H2O(l) → H2(g) + ½ O2(g) ΔH = y C3H8(g) → 3C(s) + 4 H2(g) ΔH = z Find the enthalpy change for C3H8(g) + 5 O2(g)→ 3 CO2(g) + 4 H2O(l) in terms of the defined variables. To solve this, we need to manipulate the given reactions so that when they are added together, we get our target reaction. The first step in doing this is to flip the middle reaction—the H2O (l) has to be on the right side when we finish. While we’re at it, we’ll also multiply it by four so the coefficient on the H2O (l) is correct. That gives us: 4H2(g) + 2O2(g) → 4 H2O(l) ΔH = y × –4 = –4y We also need to multiple the top reaction by 3, so the coefficient on the CO2(g) is correct. 3C(s) + 3O2(g) → 3CO2(g) ΔH = x × 3 = 3x If we add all three equations together: 4H2(g) + 2O2(g) + 3C(s) + 3O2(g) + C3H8(g) → 4H2O (l) + 3CO2(g) + 3C(s) + 4H2(g) Anything that appears on both sides of the arrow cancels out and any identical substances on the same side of the arrow combine, leaving us with our target